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Scott continuous

Let P1,P2subscriptP1subscriptP2P_{1},P_{2} be two dcpos. A functionf:P1→P2normal-:fnormal-→subscriptP1subscriptP2f:P_{1}\to P_{2} is said to be Scott continuous if for any directed setD⊆P1DsubscriptP1D\subseteq P_{1}, f⁢(⋁D)=⋁f⁢(D)fDfDf(\bigvee D)=\bigvee f(D).

First, observe that fff is monotone. If a≤baba\leq b, then f⁢(b)=f⁢(⋁{a,b})=⋁{f⁢(a),f⁢(b)}fbfabfafbf(b)=f(\bigvee\{a,b\})=\bigvee\{f(a),f(b)\}, so that f⁢(a)≤f⁢(b)fafbf(a)\leq f(b). As a result, if DDD is directed, so is f⁢(D)fDf(D).

Proposition 1.

f:P1→P2normal-:fnormal-→subscriptP1subscriptP2f:P_{1}\to P_{2} is Scott continuous iff it is continuous when P1subscriptP1P_{1} and P2subscriptP2P_{2} are equipped with the Scott topologies.

Proof.

Suppose first that fff is Scott continuous. Take an open setU∈P2UsubscriptP2U\in P_{2}. We want to show that V:=f-1⁢(U)assignVsuperscriptf1UV:=f^{{-1}}(U) is open in P1subscriptP1P_{1}. In other words, VVV is upper and that VVV has non-empty intersection with any directed set D∈P1DsubscriptP1D\in P_{1} whenever its supremum⋁DD\bigvee D lies in VVV. If a∈↑Vfragmentsanormal-↑Va\in\uparrow\!\!V, then some b∈VbVb\in V with b≤abab\leq a, which implies f⁢(b)≤f⁢(a)fbfaf(b)\leq f(a). Since f⁢(b)∈UfbUf(b)\in U, f(a)∈↑U=Ufragmentsffragmentsnormal-(anormal-)normal-↑UUf(a)\in\uparrow\!\!U=U, so a∈f-1⁢(U)=Vasuperscriptf1UVa\in f^{{-1}}(U)=V, VVV is upper. Now, suppose ⋁D∈VDV\bigvee D\in V. So ⋁f⁢(D)=f⁢(⋁D)∈UfDfDU\bigvee f(D)=f(\bigvee D)\in U. Since f⁢(D)fDf(D) is directed, there is y∈f⁢(D)∩UyfDUy\in f(D)\cap U, which means there is x∈P1xsubscriptP1x\in P_{1} such that f⁢(x)=yfxyf(x)=y and x∈D∩VxDVx\in D\cap V. This shows that VVV is Scott open.

Conversely, suppose fff is continuous (inverse of a Scott open set is Scott open). Let DDD be a directed subset of P1subscriptP1P_{1} and let d=⋁DdDd=\bigvee D. We want to show that f⁢(d)=⋁f⁢(D)fdfDf(d)=\bigvee f(D). First, for any e∈DeDe\in D, we have that e≤dede\leq d so that f⁢(e)≤f⁢(d)fefdf(e)\leq f(d) since fff is monotone. This shows ⋁f⁢(D)≤f⁢(d)fDfd\bigvee f(D)\leq f(d). Now suppose rrr is any upper bound of f⁢(D)fDf(D). We want to show that f⁢(d)≤rfdrf(d)\leq r, or f(d)∈↓rfragmentsffragmentsnormal-(dnormal-)normal-↓rf(d)\in\downarrow\!\!r. Assume not. Then f⁢(d)fdf(d) lies in U:=P2-↓rassignUlimit-fromsubscriptP2normal-↓rU:=P_{2}-\downarrow\!\!r, a Scott open set. So ⋁D=d∈f-1⁢(U)Ddsuperscriptf1U\bigvee D=d\in f^{{-1}}(U), also Scott open, which implies some e∈DeDe\in D with e∈f-1⁢(U)esuperscriptf1Ue\in f^{{-1}}(U), or f⁢(e)∈UfeUf(e)\in U. This means f⁢(e)≰rnot-less-than-or-equalsferf(e)\not\leq r, a contradiction. Thus f⁢(d)≤rfdrf(d)\leq r, and the proof is complete.
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